Mixing
Odds of picking matching 2 digits (each digit 0-9) picked randomly, order doesn't matter
up vote
3
down vote
favorite
So someone randomly picks two digits, each 0-9, so for example 23, 05, 50, 99, 00, etc.
So, if I pick 15, and the random selection is 15 or 51, I win.
I calculated the odds of winning as 3%. There are 100 possibilities (10*10). 10 of these possibilities (e.g., 55, 66) have no "counterpart" to help you win like how, say, 31 has 13. The odds of a random selection hitting a double number like 55 or 66 are 1/10 (10 out of 100 possibilities). In that world, you have a 1/10 chance of winning. You have a 90% chance of not hitting a double number and in this world you have a 1/45 chance of winning (90 remaining possibilities / 2 since you have mirror image helpers).
(1/45) * .9 + (1/10)*(1/10) = 3%
But when I run a model in excel with 500,000 rows, with random number generators I'm getting something closer to 1.9%. Is the flaw in my math logic or my excel model logic? I have a feeling I'm not thinking about something quite right in my math logic.
probability
asked 2 days ago
GLearner
161
New contributor
GLearner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
|
show 2 more comments
up vote
3
down vote
favorite
So someone randomly picks two digits, each 0-9, so for example 23, 05, 50, 99, 00, etc.
So, if I pick 15, and the random selection is 15 or 51, I win.
I calculated the odds of winning as 3%. There are 100 possibilities (10*10). 10 of these possibilities (e.g., 55, 66) have no "counterpart" to help you win like how, say, 31 has 13. The odds of a random selection hitting a double number like 55 or 66 are 1/10 (10 out of 100 possibilities). In that world, you have a 1/10 chance of winning. You have a 90% chance of not hitting a double number and in this world you have a 1/45 chance of winning (90 remaining possibilities / 2 since you have mirror image helpers).
(1/45) * .9 + (1/10)*(1/10) = 3%
But when I run a model in excel with 500,000 rows, with random number generators I'm getting something closer to 1.9%. Is the flaw in my math logic or my excel model logic? I have a feeling I'm not thinking about something quite right in my math logic.
probability
asked 2 days ago
GLearner
161
New contributor
GLearner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
How are the winning numbers randomly picked? Does the selector choose one digit randomly and then the next randomly? Or does he select one pair of numbers randomly from the list of possible winning pairs?
– Joey Kilpatrick
2 days ago
In my excel model, the numbers are independently randomly chosen. There are 4 columns each with a random number generator between 0-9 going down for 500,000 rows. The first two columns for each row are together are taken as the first number and the second two columns as the second number going down. There's a "win" when they match per the above.
– GLearner
2 days ago
See my answer for the issue
– Joey Kilpatrick
2 days ago
Bear in mind not all pairs are equally likely to be picked. $(3,3)$ will only be picked $1$ of $100$ times but $(3,7)$ will be picked $1$ is $50$. You have a $frac 1{10}$ possibility of picking a $(a,a)$ and there is a $frac 1{100}$ chance of that winning and you have a $frac 9{10}$ possibility of picking a $(a,bne a)$ and there is a $frac 1{50}$ chance of that winning. So your probability is $frac 1{10}frac 1{100} + frac 9{10}frac 1{50}= frac {19}{100}$ by my calculations.
– fleablood
2 days ago
You are confusing that you being in a world were you picked a matching pair means that the lottery master is also in the world where she picks a matching pair. She isn't. She is could still pick any of 100 ordered pairs and the prob it is yours is $frac 1{100}$. Likewise if you are in the world where you didn't, she isn't in that wolrld and her prob of matching yours is $frac 1 {50}$.
– fleablood
2 days ago
|
show 2 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
So someone randomly picks two digits, each 0-9, so for example 23, 05, 50, 99, 00, etc.
So, if I pick 15, and the random selection is 15 or 51, I win.
I calculated the odds of winning as 3%. There are 100 possibilities (10*10). 10 of these possibilities (e.g., 55, 66) have no "counterpart" to help you win like how, say, 31 has 13. The odds of a random selection hitting a double number like 55 or 66 are 1/10 (10 out of 100 possibilities). In that world, you have a 1/10 chance of winning. You have a 90% chance of not hitting a double number and in this world you have a 1/45 chance of winning (90 remaining possibilities / 2 since you have mirror image helpers).
(1/45) * .9 + (1/10)*(1/10) = 3%
But when I run a model in excel with 500,000 rows, with random number generators I'm getting something closer to 1.9%. Is the flaw in my math logic or my excel model logic? I have a feeling I'm not thinking about something quite right in my math logic.
probability
asked 2 days ago
GLearner
161
New contributor
GLearner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
So someone randomly picks two digits, each 0-9, so for example 23, 05, 50, 99, 00, etc.
So, if I pick 15, and the random selection is 15 or 51, I win.
I calculated the odds of winning as 3%. There are 100 possibilities (10*10). 10 of these possibilities (e.g., 55, 66) have no "counterpart" to help you win like how, say, 31 has 13. The odds of a random selection hitting a double number like 55 or 66 are 1/10 (10 out of 100 possibilities). In that world, you have a 1/10 chance of winning. You have a 90% chance of not hitting a double number and in this world you have a 1/45 chance of winning (90 remaining possibilities / 2 since you have mirror image helpers).
(1/45) * .9 + (1/10)*(1/10) = 3%
But when I run a model in excel with 500,000 rows, with random number generators I'm getting something closer to 1.9%. Is the flaw in my math logic or my excel model logic? I have a feeling I'm not thinking about something quite right in my math logic.
probability
probability
asked 2 days ago
GLearner
161
New contributor
GLearner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
GLearner
161
New contributor
GLearner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
GLearner
161
New contributor
GLearner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
GLearner
161
asked 2 days ago
GLearner
161
161
New contributor
GLearner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
GLearner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
GLearner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
How are the winning numbers randomly picked? Does the selector choose one digit randomly and then the next randomly? Or does he select one pair of numbers randomly from the list of possible winning pairs?
– Joey Kilpatrick
2 days ago
In my excel model, the numbers are independently randomly chosen. There are 4 columns each with a random number generator between 0-9 going down for 500,000 rows. The first two columns for each row are together are taken as the first number and the second two columns as the second number going down. There's a "win" when they match per the above.
– GLearner
2 days ago
See my answer for the issue
– Joey Kilpatrick
2 days ago
Bear in mind not all pairs are equally likely to be picked. $(3,3)$ will only be picked $1$ of $100$ times but $(3,7)$ will be picked $1$ is $50$. You have a $frac 1{10}$ possibility of picking a $(a,a)$ and there is a $frac 1{100}$ chance of that winning and you have a $frac 9{10}$ possibility of picking a $(a,bne a)$ and there is a $frac 1{50}$ chance of that winning. So your probability is $frac 1{10}frac 1{100} + frac 9{10}frac 1{50}= frac {19}{100}$ by my calculations.
– fleablood
2 days ago
You are confusing that you being in a world were you picked a matching pair means that the lottery master is also in the world where she picks a matching pair. She isn't. She is could still pick any of 100 ordered pairs and the prob it is yours is $frac 1{100}$. Likewise if you are in the world where you didn't, she isn't in that wolrld and her prob of matching yours is $frac 1 {50}$.
– fleablood
2 days ago
|
show 2 more comments
How are the winning numbers randomly picked? Does the selector choose one digit randomly and then the next randomly? Or does he select one pair of numbers randomly from the list of possible winning pairs?
– Joey Kilpatrick
2 days ago
In my excel model, the numbers are independently randomly chosen. There are 4 columns each with a random number generator between 0-9 going down for 500,000 rows. The first two columns for each row are together are taken as the first number and the second two columns as the second number going down. There's a "win" when they match per the above.
– GLearner
2 days ago
See my answer for the issue
– Joey Kilpatrick
2 days ago
Bear in mind not all pairs are equally likely to be picked. $(3,3)$ will only be picked $1$ of $100$ times but $(3,7)$ will be picked $1$ is $50$. You have a $frac 1{10}$ possibility of picking a $(a,a)$ and there is a $frac 1{100}$ chance of that winning and you have a $frac 9{10}$ possibility of picking a $(a,bne a)$ and there is a $frac 1{50}$ chance of that winning. So your probability is $frac 1{10}frac 1{100} + frac 9{10}frac 1{50}= frac {19}{100}$ by my calculations.
– fleablood
2 days ago
You are confusing that you being in a world were you picked a matching pair means that the lottery master is also in the world where she picks a matching pair. She isn't. She is could still pick any of 100 ordered pairs and the prob it is yours is $frac 1{100}$. Likewise if you are in the world where you didn't, she isn't in that wolrld and her prob of matching yours is $frac 1 {50}$.
– fleablood
2 days ago
How are the winning numbers randomly picked? Does the selector choose one digit randomly and then the next randomly? Or does he select one pair of numbers randomly from the list of possible winning pairs?
– Joey Kilpatrick
2 days ago
How are the winning numbers randomly picked? Does the selector choose one digit randomly and then the next randomly? Or does he select one pair of numbers randomly from the list of possible winning pairs?
– Joey Kilpatrick
2 days ago
In my excel model, the numbers are independently randomly chosen. There are 4 columns each with a random number generator between 0-9 going down for 500,000 rows. The first two columns for each row are together are taken as the first number and the second two columns as the second number going down. There's a "win" when they match per the above.
– GLearner
2 days ago
In my excel model, the numbers are independently randomly chosen. There are 4 columns each with a random number generator between 0-9 going down for 500,000 rows. The first two columns for each row are together are taken as the first number and the second two columns as the second number going down. There's a "win" when they match per the above.
– GLearner
2 days ago
See my answer for the issue
– Joey Kilpatrick
2 days ago
See my answer for the issue
– Joey Kilpatrick
2 days ago
Bear in mind not all pairs are equally likely to be picked. $(3,3)$ will only be picked $1$ of $100$ times but $(3,7)$ will be picked $1$ is $50$. You have a $frac 1{10}$ possibility of picking a $(a,a)$ and there is a $frac 1{100}$ chance of that winning and you have a $frac 9{10}$ possibility of picking a $(a,bne a)$ and there is a $frac 1{50}$ chance of that winning. So your probability is $frac 1{10}frac 1{100} + frac 9{10}frac 1{50}= frac {19}{100}$ by my calculations.
– fleablood
2 days ago
Bear in mind not all pairs are equally likely to be picked. $(3,3)$ will only be picked $1$ of $100$ times but $(3,7)$ will be picked $1$ is $50$. You have a $frac 1{10}$ possibility of picking a $(a,a)$ and there is a $frac 1{100}$ chance of that winning and you have a $frac 9{10}$ possibility of picking a $(a,bne a)$ and there is a $frac 1{50}$ chance of that winning. So your probability is $frac 1{10}frac 1{100} + frac 9{10}frac 1{50}= frac {19}{100}$ by my calculations.
– fleablood
2 days ago
You are confusing that you being in a world were you picked a matching pair means that the lottery master is also in the world where she picks a matching pair. She isn't. She is could still pick any of 100 ordered pairs and the prob it is yours is $frac 1{100}$. Likewise if you are in the world where you didn't, she isn't in that wolrld and her prob of matching yours is $frac 1 {50}$.
– fleablood
2 days ago
You are confusing that you being in a world were you picked a matching pair means that the lottery master is also in the world where she picks a matching pair. She isn't. She is could still pick any of 100 ordered pairs and the prob it is yours is $frac 1{100}$. Likewise if you are in the world where you didn't, she isn't in that wolrld and her prob of matching yours is $frac 1 {50}$.
– fleablood
2 days ago
|
show 2 more comments
3 Answers
3
active
oldest
votes
up vote
1
down vote
The number of possible winning pairs is the "stars and bars" problem with $2$ balls and $10$ buckets. The solution is $$
binom{2+10-1}{2}=binom{11}{2}=55
$$
Therefore there are $55$ possibilities for the winning numbers. The value $frac{1}{55}$ is about $1.8%$, so your simulation seems to assume that each of the pairs is equally possible.
However, by the way you described the problem, not every pair of winning numbers has equal chance of being the winning numbers: for instance $00$ can only be the winning numbers if a $0$ is picked and then a $0$ is picked. However, $12$ can be the winning numbers if $1$ is picked and then $2$ is picked or if $2$ is picked and then $1$ is picked. Therefore, $12$ (or $21$) has twice the chance of $00$ of being the winning numbers.
answered 2 days ago
Joey Kilpatrick
1,083121
add a comment |
up vote
1
down vote
Let's say the first person picks a two-digit number at random (including numbers with a leading zero). Each number is equally likely, with a probability of $0.01$.
Then, the second person does the same. Ten of the numbers ($00$, $11$, $22$, ... $99$) will have probability $0.01$ of matching, while all of the others will have probability of $0.02$ of matching (since there are two numbers with the same set of different digits).
Adding up all of the probabilities gives $0.01(10cdot 0.01 + 90 cdot 0.02) = 0.019 = 1.9%$, which is what you got from your simulation.
answered 2 days ago
John
22.3k32347
add a comment |
up vote
1
down vote
The error is your model.
You say that if you draw a double pair you use the phrase "In that world, you have a 1/10 chance of winning. "
You seem to be assuming that if YOU drew a pair, then the other person had to ALSO draw a pair.
And when YOU don't draw a pair you say " and in this world you have a 1/45 chance of winning" implying you are assuming the other person ALSO drew a mismatch.
But the other person draw is independent of yours. There are, with respect to order, $100$ things she could have drawn and that is true no matter what you drew. So in the world where you drew a pair she has a $frac 1{100}$ probability of matching it. And in the world were you didn't draw a pair, she has a $frac 2{100}=frac 1{50}$ probability of matching it.
So the probability of a match is
$frac 1{10}frac 1{100} + frac {9}{10}frac {2}{100} = frac {19}{1000} = 1.9%$.
=====
Alternatively:
Probabilty that you both drew a pair: $frac 1{10}frac 1{10}= frac 1{100}$. If so probability they match: $frac 1{10}$.
Probability that you both drew a mismatch: $frac 9{10}frac 9{10}=frac {81}{100}$. If so probabitilty they match: $frac 1{45}$.
Probability that you drew a pair and she didn't: $frac 1{10}frac 9{10} = frac 9{100}$. If so probability they match: $0$.
Probability that you didn't draw a pair and she did: $frac 9{10}frac 1{10} =frac 9{100}$. If so probability they match: $0$.
So probability of match:
$frac 1{100}frac 1{10} + frac {81}{100}frac 1{45} + frac 9{100}cdot 0 + frac 9{100}cdot 0= frac 1{1000} + frac {81}{4500} = frac 1{1000} + frac 9{500}=1.9%$
answered 2 days ago
fleablood
65.6k22682
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The number of possible winning pairs is the "stars and bars" problem with $2$ balls and $10$ buckets. The solution is $$
binom{2+10-1}{2}=binom{11}{2}=55
$$
Therefore there are $55$ possibilities for the winning numbers. The value $frac{1}{55}$ is about $1.8%$, so your simulation seems to assume that each of the pairs is equally possible.
However, by the way you described the problem, not every pair of winning numbers has equal chance of being the winning numbers: for instance $00$ can only be the winning numbers if a $0$ is picked and then a $0$ is picked. However, $12$ can be the winning numbers if $1$ is picked and then $2$ is picked or if $2$ is picked and then $1$ is picked. Therefore, $12$ (or $21$) has twice the chance of $00$ of being the winning numbers.
answered 2 days ago
Joey Kilpatrick
1,083121
add a comment |
up vote
1
down vote
The number of possible winning pairs is the "stars and bars" problem with $2$ balls and $10$ buckets. The solution is $$
binom{2+10-1}{2}=binom{11}{2}=55
$$
Therefore there are $55$ possibilities for the winning numbers. The value $frac{1}{55}$ is about $1.8%$, so your simulation seems to assume that each of the pairs is equally possible.
However, by the way you described the problem, not every pair of winning numbers has equal chance of being the winning numbers: for instance $00$ can only be the winning numbers if a $0$ is picked and then a $0$ is picked. However, $12$ can be the winning numbers if $1$ is picked and then $2$ is picked or if $2$ is picked and then $1$ is picked. Therefore, $12$ (or $21$) has twice the chance of $00$ of being the winning numbers.
answered 2 days ago
Joey Kilpatrick
1,083121
add a comment |
up vote
1
down vote
up vote
1
down vote
The number of possible winning pairs is the "stars and bars" problem with $2$ balls and $10$ buckets. The solution is $$
binom{2+10-1}{2}=binom{11}{2}=55
$$
Therefore there are $55$ possibilities for the winning numbers. The value $frac{1}{55}$ is about $1.8%$, so your simulation seems to assume that each of the pairs is equally possible.
However, by the way you described the problem, not every pair of winning numbers has equal chance of being the winning numbers: for instance $00$ can only be the winning numbers if a $0$ is picked and then a $0$ is picked. However, $12$ can be the winning numbers if $1$ is picked and then $2$ is picked or if $2$ is picked and then $1$ is picked. Therefore, $12$ (or $21$) has twice the chance of $00$ of being the winning numbers.
answered 2 days ago
Joey Kilpatrick
1,083121
The number of possible winning pairs is the "stars and bars" problem with $2$ balls and $10$ buckets. The solution is $$
binom{2+10-1}{2}=binom{11}{2}=55
$$
Therefore there are $55$ possibilities for the winning numbers. The value $frac{1}{55}$ is about $1.8%$, so your simulation seems to assume that each of the pairs is equally possible.
However, by the way you described the problem, not every pair of winning numbers has equal chance of being the winning numbers: for instance $00$ can only be the winning numbers if a $0$ is picked and then a $0$ is picked. However, $12$ can be the winning numbers if $1$ is picked and then $2$ is picked or if $2$ is picked and then $1$ is picked. Therefore, $12$ (or $21$) has twice the chance of $00$ of being the winning numbers.
answered 2 days ago
Joey Kilpatrick
1,083121
edited 2 days ago
answered 2 days ago
Joey Kilpatrick
1,083121
answered 2 days ago
Joey Kilpatrick
1,083121
answered 2 days ago
Joey Kilpatrick
1,083121
1,083121
add a comment |
add a comment |
up vote
1
down vote
Let's say the first person picks a two-digit number at random (including numbers with a leading zero). Each number is equally likely, with a probability of $0.01$.
Then, the second person does the same. Ten of the numbers ($00$, $11$, $22$, ... $99$) will have probability $0.01$ of matching, while all of the others will have probability of $0.02$ of matching (since there are two numbers with the same set of different digits).
Adding up all of the probabilities gives $0.01(10cdot 0.01 + 90 cdot 0.02) = 0.019 = 1.9%$, which is what you got from your simulation.
answered 2 days ago
John
22.3k32347
add a comment |
up vote
1
down vote
Let's say the first person picks a two-digit number at random (including numbers with a leading zero). Each number is equally likely, with a probability of $0.01$.
Then, the second person does the same. Ten of the numbers ($00$, $11$, $22$, ... $99$) will have probability $0.01$ of matching, while all of the others will have probability of $0.02$ of matching (since there are two numbers with the same set of different digits).
Adding up all of the probabilities gives $0.01(10cdot 0.01 + 90 cdot 0.02) = 0.019 = 1.9%$, which is what you got from your simulation.
answered 2 days ago
John
22.3k32347
add a comment |
up vote
1
down vote
up vote
1
down vote
Let's say the first person picks a two-digit number at random (including numbers with a leading zero). Each number is equally likely, with a probability of $0.01$.
Then, the second person does the same. Ten of the numbers ($00$, $11$, $22$, ... $99$) will have probability $0.01$ of matching, while all of the others will have probability of $0.02$ of matching (since there are two numbers with the same set of different digits).
Adding up all of the probabilities gives $0.01(10cdot 0.01 + 90 cdot 0.02) = 0.019 = 1.9%$, which is what you got from your simulation.
answered 2 days ago
John
22.3k32347
Let's say the first person picks a two-digit number at random (including numbers with a leading zero). Each number is equally likely, with a probability of $0.01$.
Then, the second person does the same. Ten of the numbers ($00$, $11$, $22$, ... $99$) will have probability $0.01$ of matching, while all of the others will have probability of $0.02$ of matching (since there are two numbers with the same set of different digits).
Adding up all of the probabilities gives $0.01(10cdot 0.01 + 90 cdot 0.02) = 0.019 = 1.9%$, which is what you got from your simulation.
answered 2 days ago
John
22.3k32347
answered 2 days ago
John
22.3k32347
answered 2 days ago
John
22.3k32347
answered 2 days ago
John
22.3k32347
22.3k32347
add a comment |
add a comment |
up vote
1
down vote
The error is your model.
You say that if you draw a double pair you use the phrase "In that world, you have a 1/10 chance of winning. "
You seem to be assuming that if YOU drew a pair, then the other person had to ALSO draw a pair.
And when YOU don't draw a pair you say " and in this world you have a 1/45 chance of winning" implying you are assuming the other person ALSO drew a mismatch.
But the other person draw is independent of yours. There are, with respect to order, $100$ things she could have drawn and that is true no matter what you drew. So in the world where you drew a pair she has a $frac 1{100}$ probability of matching it. And in the world were you didn't draw a pair, she has a $frac 2{100}=frac 1{50}$ probability of matching it.
So the probability of a match is
$frac 1{10}frac 1{100} + frac {9}{10}frac {2}{100} = frac {19}{1000} = 1.9%$.
=====
Alternatively:
Probabilty that you both drew a pair: $frac 1{10}frac 1{10}= frac 1{100}$. If so probability they match: $frac 1{10}$.
Probability that you both drew a mismatch: $frac 9{10}frac 9{10}=frac {81}{100}$. If so probabitilty they match: $frac 1{45}$.
Probability that you drew a pair and she didn't: $frac 1{10}frac 9{10} = frac 9{100}$. If so probability they match: $0$.
Probability that you didn't draw a pair and she did: $frac 9{10}frac 1{10} =frac 9{100}$. If so probability they match: $0$.
So probability of match:
$frac 1{100}frac 1{10} + frac {81}{100}frac 1{45} + frac 9{100}cdot 0 + frac 9{100}cdot 0= frac 1{1000} + frac {81}{4500} = frac 1{1000} + frac 9{500}=1.9%$
answered 2 days ago
fleablood
65.6k22682
add a comment |
up vote
1
down vote
The error is your model.
You say that if you draw a double pair you use the phrase "In that world, you have a 1/10 chance of winning. "
You seem to be assuming that if YOU drew a pair, then the other person had to ALSO draw a pair.
And when YOU don't draw a pair you say " and in this world you have a 1/45 chance of winning" implying you are assuming the other person ALSO drew a mismatch.
But the other person draw is independent of yours. There are, with respect to order, $100$ things she could have drawn and that is true no matter what you drew. So in the world where you drew a pair she has a $frac 1{100}$ probability of matching it. And in the world were you didn't draw a pair, she has a $frac 2{100}=frac 1{50}$ probability of matching it.
So the probability of a match is
$frac 1{10}frac 1{100} + frac {9}{10}frac {2}{100} = frac {19}{1000} = 1.9%$.
=====
Alternatively:
Probabilty that you both drew a pair: $frac 1{10}frac 1{10}= frac 1{100}$. If so probability they match: $frac 1{10}$.
Probability that you both drew a mismatch: $frac 9{10}frac 9{10}=frac {81}{100}$. If so probabitilty they match: $frac 1{45}$.
Probability that you drew a pair and she didn't: $frac 1{10}frac 9{10} = frac 9{100}$. If so probability they match: $0$.
Probability that you didn't draw a pair and she did: $frac 9{10}frac 1{10} =frac 9{100}$. If so probability they match: $0$.
So probability of match:
$frac 1{100}frac 1{10} + frac {81}{100}frac 1{45} + frac 9{100}cdot 0 + frac 9{100}cdot 0= frac 1{1000} + frac {81}{4500} = frac 1{1000} + frac 9{500}=1.9%$
answered 2 days ago
fleablood
65.6k22682
add a comment |
up vote
1
down vote
up vote
1
down vote
The error is your model.
You say that if you draw a double pair you use the phrase "In that world, you have a 1/10 chance of winning. "
You seem to be assuming that if YOU drew a pair, then the other person had to ALSO draw a pair.
And when YOU don't draw a pair you say " and in this world you have a 1/45 chance of winning" implying you are assuming the other person ALSO drew a mismatch.
But the other person draw is independent of yours. There are, with respect to order, $100$ things she could have drawn and that is true no matter what you drew. So in the world where you drew a pair she has a $frac 1{100}$ probability of matching it. And in the world were you didn't draw a pair, she has a $frac 2{100}=frac 1{50}$ probability of matching it.
So the probability of a match is
$frac 1{10}frac 1{100} + frac {9}{10}frac {2}{100} = frac {19}{1000} = 1.9%$.
=====
Alternatively:
Probabilty that you both drew a pair: $frac 1{10}frac 1{10}= frac 1{100}$. If so probability they match: $frac 1{10}$.
Probability that you both drew a mismatch: $frac 9{10}frac 9{10}=frac {81}{100}$. If so probabitilty they match: $frac 1{45}$.
Probability that you drew a pair and she didn't: $frac 1{10}frac 9{10} = frac 9{100}$. If so probability they match: $0$.
Probability that you didn't draw a pair and she did: $frac 9{10}frac 1{10} =frac 9{100}$. If so probability they match: $0$.
So probability of match:
$frac 1{100}frac 1{10} + frac {81}{100}frac 1{45} + frac 9{100}cdot 0 + frac 9{100}cdot 0= frac 1{1000} + frac {81}{4500} = frac 1{1000} + frac 9{500}=1.9%$
answered 2 days ago
fleablood
65.6k22682
The error is your model.
You say that if you draw a double pair you use the phrase "In that world, you have a 1/10 chance of winning. "
You seem to be assuming that if YOU drew a pair, then the other person had to ALSO draw a pair.
And when YOU don't draw a pair you say " and in this world you have a 1/45 chance of winning" implying you are assuming the other person ALSO drew a mismatch.
But the other person draw is independent of yours. There are, with respect to order, $100$ things she could have drawn and that is true no matter what you drew. So in the world where you drew a pair she has a $frac 1{100}$ probability of matching it. And in the world were you didn't draw a pair, she has a $frac 2{100}=frac 1{50}$ probability of matching it.
So the probability of a match is
$frac 1{10}frac 1{100} + frac {9}{10}frac {2}{100} = frac {19}{1000} = 1.9%$.
=====
Alternatively:
Probabilty that you both drew a pair: $frac 1{10}frac 1{10}= frac 1{100}$. If so probability they match: $frac 1{10}$.
Probability that you both drew a mismatch: $frac 9{10}frac 9{10}=frac {81}{100}$. If so probabitilty they match: $frac 1{45}$.
Probability that you drew a pair and she didn't: $frac 1{10}frac 9{10} = frac 9{100}$. If so probability they match: $0$.
Probability that you didn't draw a pair and she did: $frac 9{10}frac 1{10} =frac 9{100}$. If so probability they match: $0$.
So probability of match:
$frac 1{100}frac 1{10} + frac {81}{100}frac 1{45} + frac 9{100}cdot 0 + frac 9{100}cdot 0= frac 1{1000} + frac {81}{4500} = frac 1{1000} + frac 9{500}=1.9%$
answered 2 days ago
fleablood
65.6k22682
edited 2 days ago
answered 2 days ago
fleablood
65.6k22682
answered 2 days ago
fleablood
65.6k22682
answered 2 days ago
fleablood
65.6k22682
65.6k22682
add a comment |
add a comment |
GLearner is a new contributor. Be nice, and check out our Code of Conduct.
GLearner is a new contributor. Be nice, and check out our Code of Conduct.
GLearner is a new contributor. Be nice, and check out our Code of Conduct.
GLearner is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005495%2fodds-of-picking-matching-2-digits-each-digit-0-9-picked-randomly-order-doesn%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
How are the winning numbers randomly picked? Does the selector choose one digit randomly and then the next randomly? Or does he select one pair of numbers randomly from the list of possible winning pairs?
– Joey Kilpatrick
2 days ago
In my excel model, the numbers are independently randomly chosen. There are 4 columns each with a random number generator between 0-9 going down for 500,000 rows. The first two columns for each row are together are taken as the first number and the second two columns as the second number going down. There's a "win" when they match per the above.
– GLearner
2 days ago
See my answer for the issue
– Joey Kilpatrick
2 days ago
Bear in mind not all pairs are equally likely to be picked. $(3,3)$ will only be picked $1$ of $100$ times but $(3,7)$ will be picked $1$ is $50$. You have a $frac 1{10}$ possibility of picking a $(a,a)$ and there is a $frac 1{100}$ chance of that winning and you have a $frac 9{10}$ possibility of picking a $(a,bne a)$ and there is a $frac 1{50}$ chance of that winning. So your probability is $frac 1{10}frac 1{100} + frac 9{10}frac 1{50}= frac {19}{100}$ by my calculations.
– fleablood
2 days ago
You are confusing that you being in a world were you picked a matching pair means that the lottery master is also in the world where she picks a matching pair. She isn't. She is could still pick any of 100 ordered pairs and the prob it is yours is $frac 1{100}$. Likewise if you are in the world where you didn't, she isn't in that wolrld and her prob of matching yours is $frac 1 {50}$.
– fleablood
2 days ago